For the cell,$Zn_{(s)} | Zn^{2+} (1 \ M) || Ag^{+} (1 \ M) | Ag_{(s)}$. If the concentration of $Zn^{2+}$ decreases to $0.1 \ M$ at $298 \ K$,then the $EMF$ of the cell:

What will be the electromotive force of the following cell at $298 \ K$ ............. $V$
$Pt \, | \, Br_{2(\ell), 0.1M} \, | \, Br^{-}_{(aq), 0.1M} \, || \, H^{+}_{(aq), 0.1M} \, | \, H_{2(g), 1atm} \, | \, Pt$
Given: $E^{0}_{Br^{-}/Br_{2}} = -1.06 \ V$

Which of the following is correct as a Nernst equation for the given electrochemical cell ?
$Mg_{(s)}|Mg_{(aq)}^{2+}(0.1 \ M)||Cl_{(aq)}^{-}(0.1 \ M)|Cl_{2_{(g)}}(1 \ bar)|Pt_{(s)}$

Consider the cell at $25^{\circ} C$:
$Zn | Zn^{2+}_{(aq)} (1 \ M) || Fe^{3+}_{(aq)}, Fe^{2+}_{(aq)} | Pt_{(s)}$
The fraction of total iron present as $Fe^{3+}$ ion at the cell potential of $1.500 \ V$ is $X \times 10^{-2}$. The value of $X$ is $.....$ (Nearest integer).
(Given $E^{0}_{Fe^{3+} / Fe^{2+}} = 0.77 \ V, E^{0}_{Zn^{2+} / Zn} = -0.76 \ V$)

Calculate the $emf$ of the cell: $Cr | Cr^{3+}(0.1 \ M) || Fe^{2+}(0.01 \ M) | Fe$. Given: $E^0_{Cr^{3+}/Cr} = -0.75 \ V$; $E^0_{Fe^{2+}/Fe} = -0.45 \ V$. Cell reaction: $2 \ Cr_{(s)} + 3 \ Fe^{2+}_{(aq)} \rightarrow 2 \ Cr^{3+}_{(aq)} + 3 \ Fe_{(s)}$.

$2Ag^{+}_{(aq)} + Cu_{(s)} \longrightarrow Cu^{2+}_{(aq)} + 2Ag_{(s)}$
The standard potential for this reaction is $0.46 \ V$. Which change will increase the potential the most?